Haskell：leafCount 函数使用 fold fun 获取 Tree 输出 Int

Question

I define Tree type as:

data Tree a = Leaf | Node a (Tree a) (Tree a)
        deriving (Show, Eq)

Here are a fold function : How to write a function of type a-> b -> b -> b for folding a tree , basically same as what I using.

Now I want to write a function leafCount :: Tree a -> Integer , using fold and at most one helper function, I think I need to distinct leaf and node in different situation but I'm struggle with doing this, here is my code for now:

leafCount = fold sum (Node a left right)
            where 
            sum left right elem = leafCount left + leafCount right + 1

There are lot of errors in this code now that I cannot figure out at all. Please give me the basic idea and the code that can improve mine.

Answer 1

Your immediate problem is that you are doing the pattern matching on the wrong side of the = :

leafCount (Node a left right) = fold sum left right a
       where sum l r elem = leafCount left + leafCount right + 1

Answer 2

It's really easy actually, I forget to add the base case of the fold function when using it!

After I reading the fold stuff and ask other person, I figure this out!